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# help

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For a real number $$x,$$ find the minimum value of $$x^4 - 2x^2.$$

May 20, 2019

#1
+3

Let    y   =   x4  -  2x2

$$y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x$$

Now let's find what values of  x  make  $$\frac{dy}{dx}$$  be  0

$$0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1)$$

Set each factor equal to zero and solve for  x

$$\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}$$

These are the  x  values of all the critical points.

We can look at a graph to determine which is the minimum:

https://www.desmos.com/calculator/vvl7egtes1

We can see that the minumum occurs when  x  =  1  and when  x  =  -1

When  x  =  ±1 ,    x2  =  1

And when  x2  =  1 ,

y   =   x4  -  2x2   =   (x2)2  -  2x2   =   (1)2 - 2(1)   =   -1

So the minumum value of  y  is  -1

May 20, 2019

#1
+3

Let    y   =   x4  -  2x2

$$y\,=\,x^4-2x^2\\~\\ \frac{dy}{dx}\,=\,\frac{d}{dx}x^4-\frac{d}{dx}2x^2\\~\\ \frac{dy}{dx}\,=\,4x^3-4x$$

Now let's find what values of  x  make  $$\frac{dy}{dx}$$  be  0

$$0\,=\,4x^3-4x\\~\\ 0\,=\,4x(x^2-1)$$

Set each factor equal to zero and solve for  x

$$\begin{array} {lcl} 4x=0&\qquad\text{or}\qquad &x^2-1=0\\~\\ x=0&\text{or}&x^2=1\\~\\ &&x=1\quad\text{or}\quad x=-1 \end{array}$$

These are the  x  values of all the critical points.

We can look at a graph to determine which is the minimum:

https://www.desmos.com/calculator/vvl7egtes1

We can see that the minumum occurs when  x  =  1  and when  x  =  -1

When  x  =  ±1 ,    x2  =  1

And when  x2  =  1 ,

y   =   x4  -  2x2   =   (x2)2  -  2x2   =   (1)2 - 2(1)   =   -1

So the minumum value of  y  is  -1

hectictar May 20, 2019